A rod of length 50 cm is pivoted... | Rotational Motion

JEE MAIN 2019

09-01-19 S2

Question

A rod of length 50 cm is pivoted at one end. It is raised such that it makes an angle of $30^{\circ}$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s-1) will be ( $g=10 \mathrm{ms}^{-2}$ )

Select the correct option:

$\frac{\sqrt{20}}{3}$

$\sqrt{30}$ (

$\sqrt{\frac{30}{2}}$

$\frac{\sqrt{30}}{2}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Conservation of mechanical energy $$ \begin{aligned} & \mathrm{mg} \frac{\mathrm{l}}{2} \sin 30^{\circ}=\frac{1}{2} \frac{\mathrm{ml}^2}{3} \cdot \omega^2 \\ \Rightarrow & \omega^2=\frac{3 \mathrm{~g}}{2 \mathrm{l}}=\frac{30}{1} \\ \Rightarrow & \omega=\sqrt{30} \mathrm{rads} / \mathrm{s} \end{aligned} $$

Question Tags

JEE Main

Physics

Hard

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